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I am struggling with the following assignment, how should I approach this?

The offspring distribution follows: $p_0 + p_1 = 1$.

Calculate $q_1 = \mathbf{P}(X_1=0), q_1 = \mathbf{P}(X_1\neq0, X_2 = 0) , q_n = \mathbf{P}(X_1\neq0, X_{n-1} \neq 0, X_{n} = 0)$.

Find the probability of extinction for $n \geq 1$.

J.G.
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student_from_eindhoven
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2 Answers2

1

$P(X_1\neq 0, X_2\neq 0,...,x_{n-1} \neq 0,X_n=0)=p_1^{n-1} p_0$ (assuming that the initial population is of size $1$).

Kavi Rama Murthy
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1

Assuming $X_0=1$, by independence we may directly compute \begin{align} \mathbb P\left(\{X_n=0\}\cap\bigcap_{i=1}^{n-1}\{X_i\ne0\}\right) &= \mathbb P(X_n=0)\mathbb P\left(\bigcap_{i=1}^{n-1}\{X_i\ne0\}\right) \\ &= \mathbb P(X_n=0)\prod_{i=1}^{n-1}\mathbb P(X_i\ne 0)\\ &= \mathbb P(X_n=0)\prod_{i=1}^{n-1}\mathbb P(X_1\ne 0)\\ &= p_0p_1^{n-1}. \end{align} The generating function for the offspring distribution is $G(s) = p_0 + p_1s$. Let $Z_0=1$ and $Z_n = \sum_{i=1}^{Z_{n-1}}X_{n,i}$ where $X_{n,i}$ are i.i.d. with the offspring distribution. The limiting extinction probability $p^*=\mathbb P\left(\bigcup_{n=1}^\infty \{Z_n=0\}\right)$ satisfies $p^* = G(p^*)$, so $$ p^* = p_0+p_1p^* \implies p^* = \frac{p_0}{1-p_1} = 1. $$ Note that this also follows from $\mathbb E[Z_1] = p_1<1$. For each $n=1,2,\ldots$ the distribution of $Z_n$ is given by the $n$-fold convolution of the offspring distribution, i.e. $Z_n$ has generating function $G_n(s)$ where $G_1(s) = G(s)$ and $G_{n+1}(s) = G(G_n(s))$. We have $G_1(s) = \sum_{k=0}^{1-1}p_0p_1^k + p^1s$, and assuming that $G_n(s) = \sum_{k=0}^{n-1}p_0p_1^k + p_1^ns$ for some $n\geqslant 1$ we have $$ G_{n+1}(s) = G(G_n(s)) = \sum_{k=0}^{n-1}p_0p_1^k + p_1^n(p_0+p_1s) = \sum_{k=0}^n p_0p_1^k + p_1^{n+1}s, $$ so it follows by induction that $G_n(s) = \sum_{k=0}^{n-1}p_0p_1^k + p_1^ns$. Hence the probability of extinction at time $n$ is $$ \mathbb P(Z_n=0) = G_n(0) = \sum_{k=0}^{n-1}p_0p_1^k. $$ Note that $$ \lim_{n\to\infty} \mathbb P(Z_n=0) = \lim_{n\to\infty}\sum_{k=0}^{n-1}p_0p_1^k = p_0\sum_{k=0}^\infty p_1^k = \frac{p_0}{1-p_1} = 1, $$ as expected.

Math1000
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